Answer:
See explaination and attachment
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Have the solution as an attachment.
Answer: a) 0.24E+20 m-3. b) p-type extrinsic.
Explanation:
The current density in a semiconductor is composed by two types of charge carriers: electrons and holes.
This parameter, is proportional to the Electric field within the semiconductor, being the proportionality constant, the electrical conductivity of the material, that takes into account the charge carrier concentrations, and the mobility for each type.
The expression for electrical conductivity is as follows:
σ = q . ne . µe + q . np . µp
Replacing by the given values, and the value of q (charge of an electron), we can get the only unknown that remains, ne , as follows:
ne =( σ – (q . np . µp)) / q µe = (13 (Ω.m)-1 – (1.6E-19) coul(4.0E+20) m-3.0.18) m2/V-s /( (1.6E-19).0.38) coul.m2/V-s
ne = 0.24E+20.
As ne is smaller than np, this means that the semiconductor behaves like a p-type extrinsic one.
So people can tell what your think
